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Sagot :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}}x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-1}\implies 3}}[/tex]
so we're really looking for the equation of a line whose slope is 3 and passes through (1 , 10)
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{3}(x-\stackrel{x_1}{1}) \\\\\\ y-10=3x-3\implies y=3x+7[/tex]
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